Đáp án:
1,
a, \({m_{FeC{l_3}}} = 19,5g\)
b,
\(C{\% _{HCl}}dư= \dfrac{{1,24 \times 36,5}}{{409,6}} \times 100\% = 11,05\% \)
\(C{\% _{FeC{l_3}}} = \dfrac{{19,5}}{{409,6}} \times 100\% = 4,76\% \)
2,
a, \(\% {m_{Fe}} = 77,78%\) và \(\% {m_{F{e_2}{O_3}}} = 22,22\% \)
b,\(C{\% _{FeC{l_2}}} =15,32% \) và \(C{\% _{FeC{l_3}}} =3,92% \)
Giải thích các bước giải:
1,
\(\begin{array}{l}
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,06mol\\
{n_{HCl}} = \dfrac{{400 \times 14,6}}{{100 \times 36,5}} = 1,6mol\\
\to \dfrac{{{n_{HCl}}}}{6} > {n_{F{e_2}{O_3}}}
\end{array}\)
Suy ra HCl dư
\(\begin{array}{l}
\to {n_{HCl}}dư= 1,6 - 6{n_{F{e_2}{O_3}}} = 1,24mol\\
\to {n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,12mol\\
\to {m_{FeC{l_3}}} = 19,5g
\end{array}\)
Dung dịch sau phản ứng gồm: HCl dư và \(FeC{l_3}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}} = 9,6 + 400 = 409,6g\\
\to C{\% _{HCl}}dư= \dfrac{{1,24 \times 36,5}}{{409,6}} \times 100\% = 11,05\% \\
\to C{\% _{FeC{l_3}}} = \dfrac{{19,5}}{{409,6}} \times 100\% = 4,76\%
\end{array}\)
2,
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Fe}} = 5,6g\\
\to {m_{F{e_2}{O_3}}} = 7,2 - 5,6 = 1,6g\\
\to \% {m_{Fe}} = \dfrac{{5,6}}{{7,2}} \times 100\% = 77,78\% \\
\to \% {m_{F{e_2}{O_3}}} = 22,22\% \\
\end{array}\)
Dung dịch A gồm \(FeC{l_2}\) và \(FeC{l_3}\)
\(\begin{array}{l}
\to {n_{HCl}} = 2{n_{Fe}} + 6{n_{F{e_2}{O_3}}} = 0,26mol\\
\to {m_{HCl}} = 9,49g\\
\to {m_{HCl}}{\rm{dd}} = \dfrac{{9,49 \times 100}}{{12,5}} = 75,92g\\
\to {n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
\to {n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,02mol\\
\to {m_{{\rm{dd}}}} = 7,2 + 75,92 - 0,1 \times 2 = 82,92g\\
\to C{\% _{FeC{l_2}}} = \dfrac{{0,1 \times 127}}{{82,92}} \times 100\% = 15,32\% \\
\to C{\% _{FeC{l_3}}} = \dfrac{{0,02 \times 162,5}}{{82,92}} \times 100\% = 3,92\%
\end{array}\)
